Rust 如何使用`&;返回未来的组合器;自我`
我使用futures v0.1编写了这段代码:Rust 如何使用`&;返回未来的组合器;自我`,rust,future,lifetime,Rust,Future,Lifetime,我使用futures v0.1编写了这段代码: impl ArcService for (Box<MiddleWare<Request>>, Box<ArcService>) { fn call(&self, req: Request, res: Response) -> Box<Future<Item = Response, Error = Error>> { box self.0.call(re
impl ArcService for (Box<MiddleWare<Request>>, Box<ArcService>) {
fn call(&self, req: Request, res: Response) -> Box<Future<Item = Response, Error = Error>> {
box self.0.call(req).and_then(move |req| self.1.call(req, res))
}
}
pub trait ArcService: Send + Sync {
fn call(&self, req: Request, res: Response) -> Box<Future<Item = Response, Error = Error>>;
}
pub trait MiddleWare<T>: Sync + Send {
fn call<'a>(&'a self, param: T) -> Box<Future<Item = T, Error = Error> + 'a>;
}
type MiddleWareFuture<'a, I> = Box<Future<Item = I, Error = Error> + 'a>;
impl MiddleWare<Request> for Vec<Box<MiddleWare<Request>>> {
fn call(&self, request: Request) -> MiddleWareFuture<Request> {
self.iter()
.fold(box Ok(request).into_future(), |request, middleware| {
box request.and_then(move |req| middleware.call(req))
})
}
}
pub struct ArcRouter {
routes: HashMap<Method, Box<ArcService>>,
}
// Service implementation
impl hyper::Server::Service for ArcRouter {
type Response = Response;
type Request = Request;
type Error = hyper::Error;
type Future = Box<Future<Item = Self::Response, Error = Self::Error>>;
fn call(&self, req: Request) -> Box<Future<Item = Self::Response, Error = Self::Error>> {
if let Some(routeMatch) = self.matchRoute(req.path(), req.method()) {
let mut request: ArcRequest = req.into();
request.paramsMap.insert(routeMatch.params);
let response = routeMatch.handler //handler is ArcService
.call(request, ArcResponse::new())
.map(|res| res.into());
return box response;
}
// TODO: this should be handled by a user defined 404 handler
return box Ok(Response::new().with_status(StatusCode::NotFound)).into_future();
}
}
它给出了生存期错误:
错误[E0495]:由于需求冲突,无法推断适当的生存期
-->src/main.rs:16:28
|
16 |让f=未来::ok(self).map(| ex | ex.age+1);
| ^^^^
|
注意:首先,生命周期不能超过15:5在方法体上定义的匿名生命周期#1。。。
-->src/main.rs:15:5
|
15 |/fn使|成为|未来(&self)->框{
16 | |让f=future::ok(self).map(| ex | ex.age+1);
17 | |盒::新(f)
18 | | }
| |_____^
注意:…因此表达式是可赋值的(预期和示例、发现和示例)
-->src/main.rs:16:28
|
16 |让f=未来::ok(self).map(| ex | ex.age+1);
| ^^^^
=注意:但是,生存期必须对静态生存期有效。。。
注意:…因此表达式是可赋值的(应为std::boxed::Box
如何使用&self
您返回一个未来,该未来指的是self
,如下所示:
use futures::future::{self, FutureResult}; // 0.1.28
struct Example {
age: i32,
}
impl Example {
fn make_a_future(&self) -> FutureResult<&Example, ()> {
future::ok(self)
}
}
你真正的问题是“我怎样才能对编译器撒谎并试图在我的程序中引入内存不安全?”
box隐式'static
-我看到您使用了语法Box@Shepmaster我最终需要一个静态的
盒子。那么,我不清楚你的最终目标是什么。大概你在尝试更难的主题(如多线程/异步IO。如果您包含对任何对象的引用,则无法满足的静态
界限。这与未来无关。请检查编辑。有许多方法可以解决生存期问题,但这与未来
和组合符没有真正的关系。简而言之:删除引用。移动值,克隆值,克隆Rc
/Arc
指向它们的指针,…-或者通过一切跟踪所涉及的生命周期(box谢谢您的示例,我用一个正确反映我的问题的示例更新了我的问题。
use futures::future::{self, FutureResult}; // 0.1.28
struct Example {
age: i32,
}
impl Example {
fn make_a_future(&self) -> FutureResult<&Example, ()> {
future::ok(self)
}
}
use futures::{future, Future}; // 0.1.28
struct Example {
age: i32,
}
impl Example {
fn make_a_future<'a>(&'a self) -> impl Future<Item = i32, Error = ()> + 'a {
future::ok(self).map(|ex| ex.age + 1)
}
}